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Question:

$\frac{d y}{d x}+3 y=e^{-2 x}$

Solution:

The given differential equation is $\frac{d y}{d x}+p y=Q$ (where $p=3$ and $Q=e^{-2 x}$ ).

Now, I.F $=e^{\int p d x}=e^{\int 3 d x}=e^{3 x}$.

The solution of the given differential equation is given by the relation,

$y(\mathrm{I} . \mathrm{F} .)=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F} .) d x+\mathrm{C}$

$\Rightarrow y e^{3 x}=\int\left(e^{-2 x} \times e^{3 x}\right)+\mathrm{C}$

$\Rightarrow y e^{3 x}=\int e^{x} d x+\mathrm{C}$

$\Rightarrow y e^{3 x}=e^{x}+\mathrm{C}$

$\Rightarrow y=e^{-2 x}+\mathrm{C} e^{-3 x}$

This is the required general solution of the given differential equation.

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